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I didn't understand why we can't identify the projective plane with its dual.

Let's take for example a line $L=aX+bY+cZ$ with $(a,b,c)\neq (0,0,0)$ in the projective plane $\mathbb P^2$. The dual projective plane $\mathbb P^{2v}$ is these lines, i.e., for each $L=aX+bY+cZ$ line in $\mathbb P^2$, we have a point $(a:b:c)\in \mathbb P^{2v}$.

However, the lines in $\mathbb P^2$ are in bijection with $\mathbb P^2$ itself, see:

$$L=aX+bY+cZ\text{ line in}\ \mathbb P^2 \longleftrightarrow (a:b:c)\in \mathbb P^2$$

Then we have to conclude that there is a bijection between the projective plane and its dual, is that true? I couldn't prove also using what I said above why there is a bijection between $\mathbb P^2$ and $\mathbb P^{2vv}$ which I'm sure it's true.

Thanks

user42912
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1 Answers1

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Yes, the projective plane is isomorphic to (as in, the geometry is equivalent, not just that they are in bijection) its dual. The isomorphism is not canonical, but rather depends on a choice of coordinates. One isomorphism is the one you describe: to the line $\{(x:y:z) | ax + by + cz = 0\}$ associate the point $(a:b:c)$, and conversely to the point $(x:y:z)$ associate the line $\{(a:b:c) | xa + yb + zc = 0\}$.

You may think of the projective plane as consisting of all lines through the origin in $\mathbb R^3$. Then the lines correspond to planes through the origin. The identification above corresponds to identifying a line in $\mathbb R^3$ with the orthogonal plane. Or you may think of $\mathbb R\mathbb P^2$ as the round 2-dimensional sphere with antipodal points identified, so that lines are great circles. Then the identification with the dual corresponds to identifying a great circle (say, the "equator") with its "centers" (the north and south poles).

If you prefer to think in terms of linear algebra (say because you want to work over an arbitrary field, rather than $\mathbb R$), then an identification between $\mathbb P^2$ and its dual is the same as an identification between 3-dimensional vector space $V$ and its (linear) dual $V^\vee$. The identification above corresponds to choosing a basis, and identifying it with its dual basis. In terms of bases, a good notation is to write vectors in $V$ as columns, and vectors in $V^\vee$ as rows. (Then the evaluation of a dual vector on a vector is just matrix multiplication.) The identification above is just taking the transpose of a column vector to get a row vector.

All descriptions illustrate that the identification is not preserved by arbitrary projective transformations. It breaks the symmetry group from $\mathrm{PGL}(3)$ down to $\mathrm{SO}(3)$.

Theo Johnson-Freyd
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  • I thought the isomorphism was between the projective plane and the dual of the dual of the projective plane. – user42912 Feb 01 '14 at 20:01
  • The isomorphism depends on a choice of isomorphism $V \to V^\vee$, so it is not "natural," but there's an isomorphism nevertheless. – Ted Shifrin Feb 01 '14 at 20:56
  • @user42912 Maybe I misunderstood your question — I thought you were asking to identify $\mathbb P^2$ with $(\mathbb P^2)^\vee$. There is a canonical equivalence $\mathbb P^2 \cong ((\mathbb P^2)^\vee)^\vee$, as follows: a point in $((\mathbb P^2)^\vee)^\vee$ is a line in $(\mathbb P^2)^\vee$ is a point in $\mathbb P^2$. The identification given is involutive, in that if you apply it twice you get the canonical isomorphism $\mathbb P^2 \cong ((\mathbb P^2)^\vee)^\vee$ – Theo Johnson-Freyd Feb 02 '14 at 00:44
  • @TedShifrin Exactly. – Theo Johnson-Freyd Feb 02 '14 at 00:44
  • Dear @TheoJohnson-Freyd What I know is a point in $(\mathbb P^2)^\vee$ is a line in $(\mathbb P^2)$, maybe you made a mistake? – user42912 Feb 02 '14 at 05:37
  • @user42912 Here is then the exercise. Define a geometry "$(\mathbb P^2)^\vee$" by declaring its underlying set to be the lines in $\mathbb P^2$, and that three "points" in $(\mathbb P^2)^\vee$ are in a "line" if the corresponding lines in $\mathbb P^2$ share a point. Then check that this satisfies all axioms to be a projective plane. Or: work linear algebraically, where the claim is the isomorphism of a 3-dimensional space (equipped with a basis) with its dual induces the claimed isomorphism on projective planes. – Theo Johnson-Freyd Feb 02 '14 at 18:05